If $y = \tan^{-1}\left(\frac{a \cos x - b \sin x}{b \cos x + a \sin x}\right)$,then $\frac{dy}{dx}$ is equal to

Let $f(\theta) = \sin \left(\tan^{-1} \left(\frac{\sin \theta}{\sqrt{\cos 2\theta}} \right) \right)$,where $-\frac{\pi}{4} < \theta < \frac{\pi}{4}$. Then the value of $\frac{d}{d(\tan \theta)}(f(\theta))$ is

If $y=\tan ^{-1}\left[\frac{\log \left(\frac{e}{x^2}\right)}{\log \left(ex^2\right)}\right]+\tan ^{-1}\left[\frac{3+2 \log x}{1-6 \log x}\right]$,then $\frac{d^2 y}{dx^2}=$

If $y=\tan ^{-1}\left[\frac{\sin ^3(2 x)-3 x^2 \sin (2 x)}{3 x \sin ^2(2 x)-x^3}\right]$, then $\frac{d y}{d x}=$

$\frac{d}{dx} \left( \tan^{-1} \frac{x}{\sqrt{a^2 - x^2}} \right) = $

If $f'(x) = \sin^2 x$ and $y = f \left( \frac{2x - 1}{x^2 + 1} \right)$, then $\frac{dy}{dx}$ at $x = 1$ is